Tag: quantum

Definite values and Eigenstates

A Quoran asks: Once we decide that physical states are represented as vectors in a Hilbert Space and observables as operators, can we deduce that states of definite values of an observable should be eigenstates of the corresponding operator?
 
Not in any conventional sense of the word “deduce”. (It is possible to “represent” physical states as vectors and observables as operators in all kinds of ways that have nothing to do with the representation that is useful in quantum theories.)
 
But if you are talking about the usual connection in quantum mechanics that we make between states and vectors and between observables and (self-adjoint) operators, then that includes also the condition that for the observable represented by operator [math]A[/math], the distribution of observed values, when it is measured in the state represented by vector [math]\psi[/math], is such that the probability of a measurement of [math]A[/math] being in any specified interval [math]L[/math] is given in terms of the spectral projector [math]E_{L}(A)[/math] by the inner product [math]\frac{\langle \psi |E_{L}(A) \psi\rangle}{\langle \psi |\psi\rangle}[/math] (or just [math]\langle \psi |E_{L}(A) \psi\rangle[/math] if [math]\psi[/math] is normalized with [math]\langle \psi |\psi\rangle=1[/math]).
 
And if you are requiring that we decide to adopt that representation, then it does follow (essentially immediately from that extra assumption) that a state in which the observable has a definite value is represented by an eigenvector of the corresponding operator.
 
PS The spectral projectors [math]E_{L}(A)[/math] of any observable [math]A[/math] also correspond to question-type observables giving the value 1 when [math]A[/math] is observed to give a value in [math]L[/math] and 0 otherwise, and the expectation value of any question is just the probability of its value being 1 (ie of the answer being “true”). So the assumption made above actually follows from the simpler version which just says that the expectation values of observables are always given by inner products of the form [math]\langle \psi |A \psi\rangle[/math] where [math]\psi[/math] is a corresponding normalized vector and [math]A[/math] is the corresponding observable (which includes the case of spectral questions about other observables).

Source: (1003) Alan Cooper’s answer to Once we decide that physical states are represented as vectors in a Hilbert Space and observables as operators, can we deduce that states of definite values of an observable should be eigenstates of the corresponding operator? – Quora

Purely Probabilistic Positions?

What we interpret as the locations of elementary particles can certainly be compared with the predictions of regular mechanics. And they will often be quite close, so the classical predictions are actually useful. But the pattern of (usually small) variations from those predictions, while not “purely” probabilistic, does seem to have a component which cannot be explained in terms of some more precise classical properties that we just have not been able to properly measure. So our idea of a regular particle may just be something that does not really exist and what we interpret as its position may indeed by something that has an essentially probabilistic component.

Source: (1000) Alan Cooper’s answer to Is electron’s location purely probabilistic so its mechanism can’t be compared with regular mechanics, or is it just too small and too fast that the only way we can detect its location and interaction is through probability? – Quora

Quantum “Weirdness”

So far as we can tell everything is probably quantum. But it’s just that some quantum things seem more weird to us than others.

Our minds evolved to cope with situations in which the information most relevant to our survival consists of averages over large numbers of microscopic subsystems. For dealing with such averages the “classical” models of reality that we consider natural are good enough for survival (with the advantage of not requiring too much sophistication of the instincts we follow). It is only in specific (mostly artificial) contexts that the quantum nature of reality becomes relevant; so there has been no need for our instincts to take account of that, and so those instincts tend to be based on the classical approximation – and when that approximation fails it feels to us like “weirdness”.

The weirdness stops, not when things are “not quantum”, but just when (due to the averaging business) their quantum behaviour is well approximated by the classical models that correspond to our evolved instincts.

Source: (1000) Alan Cooper’s answer to There is a lot of weirdness in quantum physics at the sub-atomic level, but why does that weirdness stop once things are not quantum? – Quora

Schrödinger’s Equation for Light

A Quora Question Asks: Is it true that there is no Schrodinger equation for light because the Schrodingers’ equation is only for massive particles, and that only Maxwell’s equations can be used for light?

My answer is that it depends on what you mean by the Schrödinger Equation.

The first time many of us see that name is in reference to the position space wave equation for a single massive particle in non-relativistic quantum mechanics (and that is indeed historically appropriate, but is not the only equation that is correctly attributed to Schrödinger). In terms of this usage, the Schrödinger Equation clearly does not apply to light because the behaviour of light is not Galilean-covariant (and, unlike particles, does not even have a useful Galilean-covariant approximation).

But the name “Schrödinger Equation” is also used for the equation that governs the time evolution of any quantum system (which is basically just a statement of the fact that any unitary representation of the one dimensional translation group has a self adjoint generator – which, in the case of time translations, we identify as corresponding to the total energy of the corresponding spacelike time slice). And in that sense, as noted by Mark John Fernee, (with some further elaboration by Diógenes Figueroa) it certainly does apply to the quantized electromagnetic field and so to light.

There is also a Schrödinger Equation in this second sense for the Quantum Field Theory of electrons and positrons; however the relativistic wave equation for a single free electron is not called a Schrödinger Equation but rather is known as the Dirac Equation instead.

For the case of a single relativistic spinless particle, there is no wave equation that is first order in time, and the best we can do is the second order Klein-Gordon Equation; but there is nonetheless a Schrödinger Equation in the second sense for the corresponding scalar quantum field theory.

There is a sense in which Maxwell’s Equations play a similar role to the Dirac and Klein-Gordon equations (and the first interpretation of Schrödinger’s) as the wave equation for a single photon, but the use of Electric and Magnetic fields to extract a probability density is complicated by having to take account of gauge invariance and polarization state.

Source: (1000) Alan Cooper’s answer to Is it true that there is no Schrodinger equation for light because the Schrodingers’ equation is only for massive particles, and that only Maxwell’s equations can be used for light? – Quora

 

Infinite Wavelength of Stationary Particle

A particle does not have a wave function with respect to itself; but for any observer, the uncertainty principle tells us that if a particle could be known to have any exact velocity (and in particular if it is known to be in the observer’s rest frame with a velocity of zero) then its position would be completely unknown – and in the case of zero velocity this would make the wave function constant. (Of course, in practice we never know either the position or the momentum exactly, and this corresponds to the mathematical fact that the constant amplitude wave-function is not normalizable.)

A typical realistic position space wave function is in the form of a wave packet which has an amplitude representing the probability density multiplied by a complex phase factor which oscillates (or more precisely rotates around the unit circle in the complex plane) at a frequency corresponding to the average observed velocity. As the velocity goes to zero, the wavelength of those phase oscillations goes to infinity and the wavefunction just looks like a bump of almost constant phase. But this infinite wavelength does not mean that the wavefunction is constant, and the shape of its amplitude envelope means that its fourier transform includes contributions from frequencies other than that corresponding to the average observed velocity (and so the momentum space wave function is also a bump with width related to that in position space by the uncertainty principle).

Source: (1000) Alan Cooper’s answer to Debroglie says wavelength is inversely proportional to velocity. But velocity is relative, and it is 0 in that particle’s resting frame of reference. So what does it mean for a particle to have infinite wavelength? – Quora

wave-particle duality 

The concept of wave-particle duality in quantum mechanics is just a way of expressing the fact that many of the physical phenomena we observe (such as the interaction of light with matter and the propagation of electrons through a crystal) are not well predicted by intuitive classical models (eg of light as waves or of electrons as discrete particles), and that in some cases it looks more as if light is made up of discrete particles and electrons propagate as waves. Quantum mechanics is a mathematical model which always predicts what we do see in all situations, with the classical wave-like and particle-like behaviour being limiting cases predicted in exactly the situations that we do see them (which for light is usually more wave-like as the scale or strength of the signals is increased and more particle-like for small scale and/or very faint signals and for matter more the other way around).

Source: (1000) Alan Cooper’s answer to What is the difference between velocity and acceleration? What is the concept of wave-particle duality in quantum mechanics? – Quora

“Basis Vector” Confusion

A Quora question asks: “The wave function is contained in a Hilbert space while its basis vectors aren’t because plane waves are not square-integrable functions. Is this true for all Hilbert spaces or only for the square-integrable sub-space?”

My response: There are a number of ways you are misusing the language here, and I thought at first that your main misunderstanding may be to think that an element of a Hilbert space has its own set of basis vectors – while in quantum theories the choice of a relevant basis is more often related to an observable than to a state. But perhaps you are not associating the “basis” with a particular wave function, and rather thinking of it as associated with the position space representation as a whole. That makes more sense so let’s go back and try to describe the situation more clearly.

A wave function is a representation of an element of the Hilbert space of quantum states by a square integrable function (but the space of such functions is actually isomorphic to the entire state space as a whole and not just a proper sub-space). There are many such representations, sometimes with associated ways of identifying bases and sometimes not. In particular for any observable with purely discrete spectrum (such as the Hamiltonian of a harmonic oscillator) there is a basis of eigenvectors, and every state is represented by a square summable sequence corresponding to its spectral decomposition. Unfortunately the position and momentum observables have pure continuous spectrum and no eigenvectors, so the corresponding representations involve elements from some larger space. The usual “position space” wave function corresponds to the spectral decomposition of the position operator, and the analogue of basis vectors are actually not functions at all but rather distributions (in particular delta functions). The plane waves on the other hand are (position-space representations of) eigenfunctions of the momentum operator (but again not eigenvectors since not in the Hilbert space).

So in the end I might answer your question by saying that there is only one Hilbert space of states but that any complete set of observables can be used to represent it in terms of square integrable functions (or sequences); and that it is only in the case of pure point spectrum that the resulting spectral decomposition can be described in terms of an actual basis, while observables with continuous spectrum generally require some kind of generalized basis involving functions or distributions that do not actually correspond to normalizable states.

Source: (1000) Alan Cooper’s answer to The wave function is contained in a Hilbert space while its basis vectors aren’t because plane waves are not square-integrable functions. Is this true for all Hilbert spaces or only for the square-integrable sub-space? – Quora

Waves vs Particles

A Quora question asks: Can we measure wave properties of particles or is my contention that ‘waves travel but particles are detected’ correct?

You are right that the actual values of the quantum “wave function” are not generally observable and that the things we can actually measure are usually more like properties of particles.

But there are some ways of getting partial information about the wave function itself. For example chemistry and molecular structure gives us a way of learning about the squared magnitude of the wave function when it is a “standing wave” and a scanning tunneling microscope even gives us a more direct picture of that. And the relationships between phases at different points sometimes lead to observable effects in solid state theory and in the Aharonov-Bohm effect.

Source: (1000) Alan Cooper’s answer to Can we measure wave properties of particles or is my contention that ‘waves travel but particles are detected’ correct? – Quora

Chemistry

The Pauli exclusion principle allows us to approximate the wave functions of valence electrons by treating the inner electrons and nucleus together as a single source of potential; and then by treating the ionic cores as fixed we can solve the Schrodinger equation for the valence electrons and calculate its lowest energy level as a function of the relative coordinates of the cores. Minimizing that function then allows us to determine the optimal bond lengths and their relative orientations.

Source: (1001) Alan Cooper’s answer to How does quantum mechanics treat atomic bonds, and what role does the Pauli exclusion principle play in this context, considering also that electrons are everywhere in space according to their wave function rather than confined to fixed orbits? – Quora

Why do people have different definitions of quantum? Is quantum mechanics a logically consistent, self-consistent theory? 

Quantum Mechanics is not a single theory. In the past there have been other attempts to describe the fundamental aspects of physics which used the word “quantum” in various different senses, but to most physicists nowadays it is a class of theories characterized by the property of having the “pure states” of an isolated system represented by rank one projectors (or equivalently rays or unit vectors) in a complex Hilbert space – and by a rule for predicting the probability distributions of outcomes for various possible experimental observations. Each such theory is internally consistent, but that doesn’t mean either that they are necessarily correct in their predictions or consistent with either one another or with other theories about the physical world.

Source: (1001) Alan Cooper’s answer to Why do people have different definitions of quantum? Is quantum mechanics a logically consistent, self-consistent theory? – Quora